Problem: What is the slope of the line tangent to $f(x) = 2x^{2}+x-4$ at $x = -1$ ?
Answer: The slope of the tangent line is $ \lim_{\Delta x \to 0} \frac{f(x + \Delta x) - f(x)}{\Delta x}$ $ = \lim_{\Delta x \to 0} \frac{(2(x+\Delta x)^{2}+x+\Delta x-4) - (2x^{2}+x-4)}{\Delta x}$ $ = \lim_{\Delta x \to 0} \frac{(2(x^{2}+2x \Delta x+\Delta x^{2})+x+\Delta x-4) - (2x^{2}+x-4)}{\Delta x}$ $ = \lim_{\Delta x \to 0} \frac{2x^{2}+4(x \Delta x)+2\Delta x^{2}+x+\Delta x-4-2x^{2}-x+4}{\Delta x}$ $ = \lim_{\Delta x \to 0} \frac{4(x \Delta x)+2\Delta x^{2}+\Delta x}{\Delta x}$ $ = \lim_{\Delta x \to 0} 4x+2(\Delta x)+1$ $ = 4x+1$ $ = (4)(-1)+1$ $ = -3$